阿多米安分解法维基百科,自由的 encyclopedia 阿多米安分解法(Adomian decomposition method,简称:ADM法),是1989年美国籍阿马尼亚数学家George Adomian创建的近似分解法,用以求解非线性偏微分方程[1][2] 将非线性偏微分方程写成如下形式: L ( u ) + R ( u ) + N L ( u ) = g ( x , t ) {\displaystyle L(u)+R(u)+NL(u)=g(x,t)} 其中 L、R为线性偏微分算子,NL为非线性项。 将反算子 L − 1 = ∫ 0 t ( ) {\displaystyle L^{-1}=\int _{0}^{t}()} . 用于上式 L − 1 L ( u ) = − L − 1 R ( u ) − L − 1 N L ( u ) + L − 1 g ( x , t ) = {\displaystyle L^{-1}L(u)=-L^{-1}R(u)-L^{-1}NL(u)+L^{-1}g(x,t)=} . 得 u ( x , t ) = u ( x , 0 ) − L − 1 N L ( u ) + L − 1 g ( x , t ) {\displaystyle u(x,t)=u(x,0)-L^{-1}NL(u)+L^{-1}g(x,t)} . 令方程的解u(x,t) 为: u = u 0 + u 1 + u 2 + u 3 + ⋯ {\displaystyle u=u_{0}+u_{1}+u_{2}+u_{3}+\cdots } 非线性项 NL(u)= A 0 + A 1 + A 2 + ⋯ {\displaystyle A_{0}+A_{1}+A_{2}+\cdots } 其中 A n = 1 n ! d n d λ n f ( u ( λ ) ) ∣ λ = 0 , {\displaystyle A_{n}={\frac {1}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} \lambda ^{n}}}f(u(\lambda ))\mid _{\lambda =0},} d n d λ n u ( λ ) ∣ λ = 0 = n ! u n {\displaystyle {\frac {\mathrm {d} ^{n}}{\mathrm {d} \lambda ^{n}}}u(\lambda )\mid _{\lambda =0}=n!u_{n}} 由此得 u ( x , t ) = u ( x , 0 ) + L − 1 g ( x , t ) {\displaystyle u(x,t)=u(x,0)+L_{-1}g(x,t)} u 1 ( x , t ) = − L − 1 R u 0 − L − 1 A 0 {\displaystyle u_{1}(x,t)=-L^{-1}Ru_{0}-L^{-1}A_{0}} u n ( x , t ) = − L − 1 R u n − 1 − L − 1 A n − 1 {\displaystyle u_{n}(x,t)=-L^{-1}Ru_{n-1}-L^{-1}A_{n-1}} 近似解= u 0 ( x , t ) + u 1 ( x , t ) + u 2 ( x , t ) + u 3 ( x , t ) + ⋯ {\displaystyle u_{0}(x,t)+u_{1}(x,t)+u_{2}(x,t)+u_{3}(x,t)+\cdots }
阿多米安分解法(Adomian decomposition method,简称:ADM法),是1989年美国籍阿马尼亚数学家George Adomian创建的近似分解法,用以求解非线性偏微分方程[1][2] 将非线性偏微分方程写成如下形式: L ( u ) + R ( u ) + N L ( u ) = g ( x , t ) {\displaystyle L(u)+R(u)+NL(u)=g(x,t)} 其中 L、R为线性偏微分算子,NL为非线性项。 将反算子 L − 1 = ∫ 0 t ( ) {\displaystyle L^{-1}=\int _{0}^{t}()} . 用于上式 L − 1 L ( u ) = − L − 1 R ( u ) − L − 1 N L ( u ) + L − 1 g ( x , t ) = {\displaystyle L^{-1}L(u)=-L^{-1}R(u)-L^{-1}NL(u)+L^{-1}g(x,t)=} . 得 u ( x , t ) = u ( x , 0 ) − L − 1 N L ( u ) + L − 1 g ( x , t ) {\displaystyle u(x,t)=u(x,0)-L^{-1}NL(u)+L^{-1}g(x,t)} . 令方程的解u(x,t) 为: u = u 0 + u 1 + u 2 + u 3 + ⋯ {\displaystyle u=u_{0}+u_{1}+u_{2}+u_{3}+\cdots } 非线性项 NL(u)= A 0 + A 1 + A 2 + ⋯ {\displaystyle A_{0}+A_{1}+A_{2}+\cdots } 其中 A n = 1 n ! d n d λ n f ( u ( λ ) ) ∣ λ = 0 , {\displaystyle A_{n}={\frac {1}{n!}}{\frac {\mathrm {d} ^{n}}{\mathrm {d} \lambda ^{n}}}f(u(\lambda ))\mid _{\lambda =0},} d n d λ n u ( λ ) ∣ λ = 0 = n ! u n {\displaystyle {\frac {\mathrm {d} ^{n}}{\mathrm {d} \lambda ^{n}}}u(\lambda )\mid _{\lambda =0}=n!u_{n}} 由此得 u ( x , t ) = u ( x , 0 ) + L − 1 g ( x , t ) {\displaystyle u(x,t)=u(x,0)+L_{-1}g(x,t)} u 1 ( x , t ) = − L − 1 R u 0 − L − 1 A 0 {\displaystyle u_{1}(x,t)=-L^{-1}Ru_{0}-L^{-1}A_{0}} u n ( x , t ) = − L − 1 R u n − 1 − L − 1 A n − 1 {\displaystyle u_{n}(x,t)=-L^{-1}Ru_{n-1}-L^{-1}A_{n-1}} 近似解= u 0 ( x , t ) + u 1 ( x , t ) + u 2 ( x , t ) + u 3 ( x , t ) + ⋯ {\displaystyle u_{0}(x,t)+u_{1}(x,t)+u_{2}(x,t)+u_{3}(x,t)+\cdots }